cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-3 of 3 results.

A006629 Self-convolution 4th power of A001764, which enumerates ternary trees.

Original entry on oeis.org

1, 4, 18, 88, 455, 2448, 13566, 76912, 444015, 2601300, 15426840, 92431584, 558685348, 3402497504, 20858916870, 128618832864, 797168807855, 4963511449260, 31032552351570, 194743066471800, 1226232861415695
Offset: 0

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Author

Keywords

Comments

Sum of root degrees of all noncrossing trees on nodes on a circle. - Emeric Deutsch

References

  • H. M. Finucan, Some decompositions of generalized Catalan numbers, pp. 275-293 of Combinatorial Mathematics IX. Proc. Ninth Australian Conference (Brisbane, August 1981). Ed. E. J. Billington, S. Oates-Williams and A. P. Street. Lecture Notes Math., 952. Springer-Verlag, 1982.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Programs

  • Magma
    A006629:= func< n | 2*Binomial(3*n+3,n)/(n+2) >;
    [A006629(n): n in [0..40]]; // G. C. Greubel, Aug 29 2025
    
  • Mathematica
    Table[2*Binomial[3*n+3,n]/(n+2), {n,0,40}] (* G. C. Greubel, Aug 29 2025 *)
  • PARI
    a(n)=my(m=4);binomial(3*n+m-1,n)*m/(2*n+m) /* 4th power of A001764 with offset n=0 */ \\ Paul D. Hanna, May 10 2008
    
  • SageMath
    def A006629(n): return 2*binomial(3*n+3,n)//(n+2)
    print([A006629(n) for n in range(41)]) # G. C. Greubel, Aug 29 2025

Formula

a(n) = 2*binomial(3*n+3,n)/(n+2). - Emeric Deutsch
a(n) = (n+1) * A000139(n+1). - F. Chapoton, Feb 23 2024
G.f.: hypergeom( [ 2, 5/3, 4/3 ]; [ 3, 5/2 ]; 27*x/4 ).
G.f.: A(x) = G(x)^4 where G(x) = 1 + x*G(x)^3 = g.f. of A001764 giving a(n)=C(3n+m-1,n)*m/(2n+m) at power m=4 with offset n=0. - Paul D. Hanna, May 10 2008
G.f.: (((4*sin(arcsin((3*sqrt(3*x))/2)/3))/(sqrt(3*x))-1)^2-1)/(4*x). - Vladimir Kruchinin, Feb 17 2023
E.g.f.: hypergeom([4/3, 5/3, 2]; [1, 5/2, 3]; 27*x/4). - G. C. Greubel, Aug 29 2025

Extensions

More precise definition from Paul D. Hanna, May 10 2008

A233657 a(n) = 10 * binomial(3*n+10,n)/(3*n+10).

Original entry on oeis.org

1, 10, 75, 510, 3325, 21252, 134550, 848250, 5340060, 33622600, 211915132, 1337675430, 8458829925, 53591180360, 340185835500, 2163581913780, 13786238414025, 88004926973250, 562763873596575, 3604713725613000, 23126371951808268, 148594788106641360
Offset: 0

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Author

Tim Fulford, Dec 14 2013

Keywords

Comments

Fuss-Catalan sequence is a(n,p,r) = r*binomial(np+r,n)/(np+r), this is the case p=3, r=10.

Crossrefs

Programs

  • Magma
    [10*Binomial(3*n+10, n)/(3*n+10): n in [0..30]];
  • Maple
    A233657:=n->10*binomial(3*n+10,n)/(3*n+10): seq(A233657(n), n=0..20); # Wesley Ivan Hurt, Oct 10 2014
  • Mathematica
    Table[10 Binomial[3 n + 10, n]/(3 n + 10), {n, 0, 30}]
  • PARI
    a(n) = 10*binomial(3*n+10,n)/(3*n+10);
    
  • PARI
    {a(n)=local(B=1); for(i=0, n, B=(1+x*B^(3/10))^10+x*O(x^n)); polcoeff(B, n)}
    

Formula

G.f. satisfies: B(x) = {1 + x*B(x)^(p/r)}^r, here p=3, r=10.
+2*n*(n+5)*(2*n+9)*a(n) -3*(3*n+7)*(n+3)*(3*n+8)*a(n-1)=0. - R. J. Mathar, Feb 16 2018
E.g.f.: F([10/3, 11/3, 4], [1, 11/2, 6], 27*x/4), where F is the generalized hypergeometric function. - Stefano Spezia, Oct 08 2019

A355172 The Fuss-Catalan triangle of order 2, read by rows. Related to ternary trees.

Original entry on oeis.org

1, 0, 1, 0, 1, 3, 0, 1, 5, 12, 0, 1, 7, 25, 55, 0, 1, 9, 42, 130, 273, 0, 1, 11, 63, 245, 700, 1428, 0, 1, 13, 88, 408, 1428, 3876, 7752, 0, 1, 15, 117, 627, 2565, 8379, 21945, 43263, 0, 1, 17, 150, 910, 4235, 15939, 49588, 126500, 246675
Offset: 0

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Author

Peter Luschny, Jun 25 2022

Keywords

Comments

The Fuss-Catalan triangle of order m is a regular, (0, 0)-based table recursively defined as follows: Set row(0) = [1] and row(1) = [0, 1]. Now assume row(n-1) already constructed and duplicate the last element of row(n-1). Next apply the cumulative sum m times to this list to get row(n). Here m = 2. (See the Python program for a reference implementation.)
This definition also includes the Fuss-Catalan numbers A001764(n) = T(n, n), or row 3 in A355262. For m = 1 see A355173 and for m = 3 A355174. More generally, for n >= 1 all Fuss-Catalan sequences (A355262(n, k), k >= 0) are the main diagonals of the Fuss-Catalan triangles of order n - 1.

Examples

			Table T(n, k) begins:
  [0] [1]
  [1] [0, 1]
  [2] [0, 1,  3]
  [3] [0, 1,  5, 12]
  [4] [0, 1,  7, 25,  55]
  [5] [0, 1,  9, 42, 130,  273]
  [6] [0, 1, 11, 63, 245,  700, 1428]
  [7] [0, 1, 13, 88, 408, 1428, 3876, 7752]
Seen as an array reading the diagonals starting from the main diagonal:
  [0] 1, 1,  3, 12,  55,  273,  1428,   7752,   43263,  246675, ...  A001764
  [1] 0, 1,  5, 25, 130,  700,  3876,  21945,  126500,  740025, ...  A102893
  [2] 0, 1,  7, 42, 245, 1428,  8379,  49588,  296010, 1781325, ...  A102594
  [3] 0, 1,  9, 63, 408, 2565, 15939,  98670,  610740, 3786588, ...  A230547
  [4] 0, 1, 11, 88, 627, 4235, 27830, 180180, 1157013, 7396972, ...
		

Crossrefs

A001764 (main diagonal), A102893 (subdiagonal), A102594 (diagonal 2), A230547 (diagonal 3), A005408 (column 2), A071355 (column 3), A006629 (row sums), A143603 (variant).
Cf. A123110 (triangle of order 0), A355173 (triangle of order 1), A355174 (triangle of order 3), A355262 (Fuss-Catalan array).

Programs

  • Python
    from functools import cache
    from itertools import accumulate
    @cache
    def Trow(n: int) -> list[int]:
        if n == 0: return [1]
        if n == 1: return [0, 1]
        row = Trow(n - 1) + [Trow(n - 1)[n - 1]]
        return list(accumulate(accumulate(row)))
    for n in range(9): print(Trow(n))

Formula

The general formula for the Fuss-Catalan triangles is, for m >= 0 and 0 <= k <= n:
FCT(n, k, m) = (m*(n - k) + m + 1)*(m*n + k - 1)!/((m*n + 1)!*(k - 1)!) for k > 0 and FCT(n, 0, m) = 0^n. The case considered here is T(n, k) = FCT(n, k, 2).
T(n, k) = (2*n - 2*k + 3)*(2*n + k - 1)!/((2*n + 1)!*(k - 1)!) for k > 0; T(n, 0) = 0^n.
The g.f. of row n of the FC-triangle of order m is 0^n + (x - (m + 1)*x^2) / (1 - x)^(m*n + 2), thus:
T(n, k) = [x^k] (0^n + (x - 3*x^2)/(1 - x)^(2*n + 2)).
Showing 1-3 of 3 results.